Question: Let $R$ be the region enclosed by the polar curve $r(\theta)=3\ln(\theta)$ where $\dfrac{\pi}{2}\leq \theta\leq \dfrac{7\pi}{8}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int^{\scriptsize\dfrac{\pi}{2}}_{\scriptsize\dfrac{7\pi}{8}}9\cdot \ln^2(\theta)\,d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{7\pi}{8}}\dfrac{9}{2}\cdot \ln^2(\theta)\,d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{7\pi}{8}}9\cdot \ln^2(\theta)\,d\theta$ (Choice D) D $ \int^{\scriptsize\dfrac{\pi}{2}}_{\scriptsize\dfrac{7\pi}{8}}\dfrac{9}{2}\cdot \ln^2(\theta)\,d\theta$
Solution: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=3\ln(\theta)}$, ${\alpha=\dfrac{\pi}{2}}$, and ${\beta=\dfrac{7\pi}{8}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{2}}}^{{\scriptsize\dfrac{7\pi}{8}}}\dfrac{1}{2}\left({3\ln(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{7\pi}{8}}\dfrac{1}{2}\cdot 9\ln^2(\theta)\,d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{7\pi}{8}}\dfrac{9}{2}\cdot \ln^2(\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{7\pi}{8}}\dfrac{9}{2}\cdot \ln^2(\theta)\,d\theta$